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2 years ago

Plisss help on my assesment

Mathematics

1 answer:

Anon25 [30]2 years ago

6 0

Answer:

The first one is equivalent [the 6x+48 = 2(3x+24)] and the second one is NOT equivalent [the 7x+21 ≠ 2(5x+3)]

Step-by-step explanation:

Just follow distributive property to solve these. You can ignore the first expression in both until you have to compare the answers.

1. 6x+48 and 2(3x+24)

2(3x+24) ---> 2(3x) + 2(24) ---> 6x + 48

Bring in the first expression ~ 6x+48 and 6x+48

They are the same, so they are equivalent

2. 7x+21 and 2(5x+3)

2(5x+3) ---> 2(5x) + 2(3) ----> 10x + 6

Bring in the first expression ~ 7x+21 and 10x + 6

They are NOT the same, so they are NOT equivalent

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Step-by-step explanation:

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2 years ago

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1 year ago

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PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%

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Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

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